SME
Linear Algebra · Problem type
An eigenvector of a 3 × 3 matrix
Use this for “λ = 3 is an eigenvalue of [[2, 0, 0], [1, 3, 0], [0, 1, 1]]. Find an eigenvector for it.”
Diagram
3 equations in 3 unknowns, the right-hand sides after the bar.
A row reads 0 = 0: it says nothing new, so there are infinitely many solutions.
The solution: x₁ = 0, x₂ = 2, x₃ = 1.
This is an example. Type your own values to start a new problem.
A
=
λ
=
,
v
=
m₁₁
M = A − λI, row 1, column 1
worked out (not typed)
m₂₂
M = A − λI, row 2, column 2
worked out (not typed)
m₃₃
M = A − λI, row 3, column 3
worked out (not typed)
det M
Determinant of M = A − λI
worked out (not typed)
n₁
Cross product of two rows, entry 1
worked out (not typed)
n₂
Cross product of two rows, entry 2
worked out (not typed)
n₃
Cross product of two rows, entry 3
worked out (not typed)
Formulas
m₁₁
=
a₁₁
−
λ
−1 = 2 − 3
m₂₂
=
a₂₂
−
λ
0 = 3 − 3
m₃₃
=
a₃₃
−
λ
−2 = 1 − 3
det
M
=
m₁₁
×
(
m₂₂
×
m₃₃
−
a₂₃
×
a₃₂
)
−
a₁₂
×
(
a₂₁
×
m₃₃
−
a₂₃
×
a₃₁
)
+
a₁₃
×
(
a₂₁
×
a₃₂
−
m₂₂
×
a₃₁
)
0 = −1 × (0 × (−2) − 0 × 1) − 0 × (1 × (−2) − 0 × 0) + 0 × (1 × 1 − 0 × 0)
n₁
=
entry
1
of
rᵢ
×
rⱼ,
two
rows
of
M
=
[[
m₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
m₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
m₃₃
]]
0 = entry 1 of rᵢ × rⱼ, two rows of M = [[−1, 0, 0], [1, 0, 0], [0, 1, −2]]
n₂
=
entry
2
of
rᵢ
×
rⱼ,
two
rows
of
M
=
[[
m₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
m₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
m₃₃
]]
−2 = entry 2 of rᵢ × rⱼ, two rows of M = [[−1, 0, 0], [1, 0, 0], [0, 1, −2]]
n₃
=
entry
3
of
rᵢ
×
rⱼ,
two
rows
of
M
=
[[
m₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
m₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
m₃₃
]]
−1 = entry 3 of rᵢ × rⱼ, two rows of M = [[−1, 0, 0], [1, 0, 0], [0, 1, −2]]
x₁
=
n₁
the last of
n₁
,
n₂
,
n₃
that is not 0
0
=
0
the last of 0, −2, −1 that is not 0
x₂
=
n₂
the last of
n₁
,
n₂
,
n₃
that is not 0
2
=
−
2
the last of 0, −2, −1 that is not 0
x₃
=
n₃
the last of
n₁
,
n₂
,
n₃
that is not 0
1
=
−
1
the last of 0, −2, −1 that is not 0
Assumptions
•
The eigenvectors for λ are the solutions of (A − λI)v = 0, so det(A − λI) must be 0.
•
Here A − λI has rank 2: one free variable, so the eigenvectors lie on one line.
•
v is orthogonal to every row of A − λI, so the cross product of two rows that are not parallel lies along v.
•
v is scaled so its last nonzero entry, the free variable, is 1.
Step-by-step
WE KNOW
a₁₁ = 2, a₁₂ = 0, a₁₃ = 0, a₂₁ = 1, a₂₂ = 3, a₂₃ = 0, a₃₁ = 0, a₃₂ = 1, a₃₃ = 1, λ = 3
FIND
M = A − λI, row 1, column 1 (m₁₁), M = A − λI, row 2, column 2 (m₂₂), M = A − λI, row 3, column 3 (m₃₃), determinant of M = A − λI (det M), cross product of two rows, entry 1 (n₁), cross product of two rows, entry 2 (n₂), cross product of two rows, entry 3 (n₃), eigenvector, entry 1 (x₁), eigenvector, entry 2 (x₂), eigenvector, entry 3 (x₃)
Step 1 · Find M = A − λI, row 1, column 1 (m₁₁)
Use m₁₁ = a₁₁ − λ
M
=
A
−
λI
is
A
with
λ
taken
off
each
entry
on
the
diagonal.
m₁₁
=
2
−
3
m₁₁
=
−1
Step 2 · Find M = A − λI, row 2, column 2 (m₂₂)
Use m₂₂ = a₂₂ − λ
Take
λ
off
the
next
diagonal
entry.
m₂₂
=
3
−
3
m₂₂
=
0
Step 3 · Find M = A − λI, row 3, column 3 (m₃₃)
Use m₃₃ = a₃₃ − λ
Take
λ
off
the
next
diagonal
entry.
m₃₃
=
1
−
3
m₃₃
=
−2
Step 4 · Find determinant of M = A − λI (det M)
Use det M = m₁₁ × (m₂₂ × m₃₃ − a₂₃ × a₃₂) − a₁₂ × (a₂₁ × m₃₃ − a₂₃ × a₃₁) + a₁₃ × (a₂₁ × a₃₂ − m₂₂ × a₃₁)
Expand
along
row
1.
It
must
be
0
for
λ
to
be
an
eigenvalue.
det M = −1 × (0 × (−2) − 0 × 1) − 0 × (1 × (−2) − 0 × 0) + 0 × (1 × 1 − 0 × 0)
det M = −1 × 0 − 0 × (−2)
det M = 0
Step 5 · Find cross product of two rows, entry 1 (n₁)
Use n₁ = entry 1 of rᵢ × rⱼ, two rows of M = [[m₁₁, a₁₂, a₁₃], [a₂₁, m₂₂, a₂₃], [a₃₁, a₃₂, m₃₃]]
The cross product of rows 1 and 3 of M is orthogonal to both rows, so it solves Mv = 0.
n₁
=
a₁₂
×
m₃₃
−
a₁₃
×
a₃₂
n₁
=
0
×
(−2)
−
0
×
1
n₁
=
0
Step 6 · Find cross product of two rows, entry 2 (n₂)
Use n₂ = entry 2 of rᵢ × rⱼ, two rows of M = [[m₁₁, a₁₂, a₁₃], [a₂₁, m₂₂, a₂₃], [a₃₁, a₃₂, m₃₃]]
The next entry of the same cross product, around the cycle.
n₂
=
a₁₃
×
a₃₁
−
m₁₁
×
m₃₃
n₂
=
0
×
0
−
(−1)
×
(−2)
n₂
=
0
−
2
n₂
=
−2
Step 7 · Find cross product of two rows, entry 3 (n₃)
Use n₃ = entry 3 of rᵢ × rⱼ, two rows of M = [[m₁₁, a₁₂, a₁₃], [a₂₁, m₂₂, a₂₃], [a₃₁, a₃₂, m₃₃]]
The next entry of the same cross product, around the cycle.
n₃
=
m₁₁
×
a₃₂
−
a₁₂
×
a₃₁
n₃
=
−1
×
1
−
0
×
0
n₃
=
−1
Step 8 · Find eigenvector, entry 1 (x₁)
Use x₁ = n₁ ÷ (the last of n₁, n₂, n₃ that is not 0)
Scale
n
so
the
free
variable
x₃
is
1:
divide
each
entry
by
it.
x₁
=
n₁
n₃
x₁
=
0
−1
x₁
=
0
Step 9 · Find eigenvector, entry 2 (x₂)
Use x₂ = n₂ ÷ (the last of n₁, n₂, n₃ that is not 0)
Divide the next entry by the same number.
x₂
=
n₂
n₃
x₂
=
−
2
−1
x₂
=
2
Step 10 · Find eigenvector, entry 3 (x₃)
Use x₃ = n₃ ÷ (the last of n₁, n₂, n₃ that is not 0)
Divide the next entry by the same number.
x₃
=
n₃
n₃
x₃
=
−
1
−1
Check: [[2, 0, 0], [1, 3, 0], [0, 1, 1]] × ⟨0, 2, 1⟩ = ⟨0, 6, 3⟩ = 3 × ⟨0, 2, 1⟩
x₃
=
1
Check
Put the numbers back in. Both sides should match.
−1 = 2 − 3 ✓
0 = 3 − 3 ✓
−2 = 1 − 3 ✓
0 = −1 × (0 × (−2) − 0 × 1) − 0 × (1 × (−2) − 0 × 0) + 0 × (1 × 1 − 0 × 0) ✓
0 = 0 × (−2) − 0 × 1 ✓
(−2) = 0 × 0 − (−1) × (−2) ✓
(−1) = (−1) × 1 − 0 × 0 ✓
0
=
0
−1
✓
2
=
−2
−1
✓
1
=
−1
−1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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